# Graham-Rothschild via Hales-Jewett

I am currently reading the recent preprint of Dodos, Kanellopoulos, Tyros, where the ambitiously short proof of Density Hales Jewett theorem is provided. The important ingredient is Graham-Rothschild theorem. The authors say that it follows from the HJ by some standard Ramsey arguments, but I can not find them myself, at least immediately. Is it written anywhere? Original paper of Graham and Rothschild looks too long for being used in "simple self-contained proof" of anything. Polymath's DHJ proof does not use GR at all, on first glance.

-
@Kristal, it is exactly the paper I mean. Now I asked its authors directly and got a satisfactory answer with references. – Fedor Petrov Oct 29 '12 at 12:36
Fedor: Could you share the answer you got? I am struggling with the same exact problem. – Joel Moreira Feb 15 '13 at 20:14
"The coloring result we use is indeed a special case of the Graham-Rothchild theorem. We do not say much about it in the paper, since this staff is considered to be part of the folklore. In any case,here are some references. 1) A simple and self-contained proof of the full Graham-Rothchild theorem can be found in: H. J. Promel and B. Voigt, Graham-Rothschild parameter sets, "Mathematics of Ramsey Theory", Springer-Verlag, Berlin (1990), 113-149.See in particular, Section 4, page 128. 2) Another excellent reference is: R. McCutcheon, Elemental Methods in Ergodic Ramsey Theory, Lect" – Fedor Petrov Feb 22 '13 at 21:33

## 1 Answer

In Randall - McCutcheon's book "Elemental methods in ergodic Ramsey theory" a stronger version of GR's theorem is prooved about block subspaces (theorem 2.4.1). The only theorems you need in order to proove it is Hales-Jewett and Folkman's theorem (about finite unions which you can proove using Hales - Jewett theorem again).

Another proof is given in Graham - Rothschild -Spencer book about Ramsey theory using HJ and linear algebra.

-
Thanks! Probably, linear algebra is used for linear algebraic version of Graham-Rothschild, while we need a combinatorial one? – Fedor Petrov Oct 15 '12 at 5:59
You're wright. If you want a proof for the combinatorial version of the thorem it is the one a mentioned in the first place. – T.Karageorgos Oct 15 '12 at 15:45